The Encyclopedia of Economic Principals

Chapter 8

Social Choice, Voting, and Impossibility

Count every pairwise contest before trusting a collective ranking.

Four of the chapter's worked examples, made interactive: an agenda that picks the winner, single-peaked preferences that restore a majority order, the median budget against challengers, and the cycle created by rights plus unanimity.

Every example here is a constructed teaching example: it uses the hypothetical numbers of the chapter's worked examples, plus a few values added for comparison and labelled as such in each panel. Nothing here measures a real market, firm or household.

Demonstration 1 of 4

The agenda picks the winner

With the same five ballots, can the order of votes decide the committee's choice?

When majority preference cycles, every plan loses to some other plan. The last plan introduced faces only the survivor, so the agenda setter chooses the outcome.

Equation, written in LaTeX: A\succ_M T\succ_M O\succ_M A.

Scroll sideways for the whole equation

Members 1 and 2 rank A > T > O; members 3 and 4 rank T > O > A; member 5's ranking is the control. In each round the survivor meets the next plan by simple majority.

Predict first. If member 5 switches to A > O > T, does the agenda order still matter?

Your prediction

Choose an example

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Figure: The agenda picks the winner. Tournament triangle of A, T and O with pairwise tallies; agenda A vs T first, then O; winner O.
First vote: A vs T, then O, Member 5 ranking: O > A > T
Constructed example: the chapter's hypothetical five-member committee and both rankings for member 5; all three agendas are from the book.

Calculated values

A vs T
3 to 2
T vs O
4 to 1
O vs A
3 to 2
First vote
A vs T: A wins
Winner
outsourcing (O)
Condorcet winner
none (cycle)

Member 5 ranks O > A > T. First A meets T: 3 + 2 = 5 votes, A gets 3 and T gets 2, so A survives. Then A meets O, 2 to 3, so O wins. Majority preference cycles, so whoever sets the agenda picks the winner.

Worked steps

  1. A vs T: 3 to 2, A survives
  2. A vs O: 2 to 3, O wins
  3. Majority preference cycles, so whoever sets the agenda picks the winner.

Use the idea

Before trusting a sequence of majority votes, check every pairwise contest for a cycle.

Where the conclusion applies

Sincere voting, strict rankings and a fixed agenda. Strategic voters may vote against their ranking in early rounds.

Check your understanding: With member 5 at A > O > T and the agenda T vs O first, who wins?
T beats O 4 to 1, then A beats T 3 to 2, so A wins; A also wins every other agenda because it beats O 3 to 2.

Chapter 8 source: section "Condorcet paradox".

Demonstration 2 of 4

Single-peaked preferences restore a majority order

When does majority rule give a coherent ranking of three transport plans?

In the cyclic profile each plan beats one rival and loses to the other. On a common spending line with single peaks, the median voter's favourite beats every alternative and the majority relation is transitive.

Equation, written in LaTeX: R\succ_M B\succ_M D\succ_M R.

Scroll sideways for the whole equation

R is rail, B bus and D road repair, ordered by spending D < B < R. A profile is single-peaked when each voter's ranking falls away on both sides of a favourite on that line.

Predict first. In the single-peaked profile, if voter 3's favourite moves to B (B > R > D), does the majority winner change?

Your prediction

Choose an example

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Figure: Single-peaked preferences restore a majority order. Rank profiles of three voters along the spending line D, B, R (cyclic profile); majority: cycle.
Profile: Cyclic, Voter 3 ranking (single-peaked profile): R > B > D
Constructed example: the chapter's hypothetical cyclic and single-peaked transport profiles; voter 3 rankings B > R > D and D > B > R are added for comparison.

Calculated values

Pairwise results
R over B 2 to 1; B over D 2 to 1; D over R 2 to 1
Single-peaked on D < B < R
no
Majority order
cycle
Majority winner
none

Each contest splits the 3 votes, for example R vs B gets 2 + 1 = 3: R over B 2 to 1; B over D 2 to 1; D over R 2 to 1. So the majority relation cycles, so there is no majority order. Voter 3's control applies to the single-peaked profile only.

Worked steps

  1. R vs B: 2 + 1 = 3 votes, 2 to 1
  2. B vs D: 2 + 1 = 3 votes, 2 to 1
  3. D vs R: 2 + 1 = 3 votes, 2 to 1
  4. Order: cycle

Use the idea

Check whether the options lie on one dimension that every voter judges the same way before relying on majority rule.

Where the conclusion applies

Three sincere voters with strict rankings. Restricting the domain to single-peaked profiles gives up Arrow's unrestricted domain condition.

Check your understanding: In the single-peaked profile with voter 3 ranking D > B > R, what is the majority order?
D over B 2 to 1 (voters 1 and 3), D over R 3 to 0, B over R 3 to 0: D > B > R.

Chapter 8 source: section "Arrow's impossibility theorem".

Demonstration 3 of 4

The median budget beats every challenger

Why does the median voter's ideal budget win every pairwise vote?

Whichever side the challenger is on, the median voter and every voter on the other side prefer 50. That is at least four of seven, a majority.

Equation, written in LaTeX: (10,25,40,50,65,80,95).

Scroll sideways for the whole equation

Seven voters' ideal library budgets in millions of dollars. Each prefers the proposal closer to their ideal, u(x) = -|x - m|. The cutline is halfway between the two proposals.

Predict first. Does a far challenger like 90 lose by more than a near one like 70?

Your prediction

Choose an example

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Figure: The median budget beats every challenger. Seven voter ideals on a budget line with the median 50, challenger 35 and the cutline at 42.5; 50 wins 4 to 3.
Challenger budget ($ million): 35
Constructed example: the chapter's hypothetical seven voters (challengers 35 and 70); challengers 20 and 90 are added for comparison.

Calculated values

Votes for 50
4
Votes for the challenger
3
Cutline
42.5
Winner
50

Voters closer to 50 than to 35 are those on 50's side of the midpoint (50 + 35) / 2 = 42.5: 50, 65, 80 and 95. So 50 wins 4 to 3. The median's coalition always includes the median voter and everyone beyond, so 50 cannot lose.

Worked steps

  1. Cutline = (50 + 35) / 2 = 42.5
  2. Prefer 50: 50, 65, 80 and 95 = 4 voters
  3. Prefer 35: 3 voters
  4. Result 4 to 3

Use the idea

With one policy dimension and single-peaked preferences, expect proposals to converge on the median voter's ideal.

Where the conclusion applies

One dimension, sincere voting, single-peaked preferences and an odd number of voters.

Check your understanding: Against a $90 million challenger, what is the tally?
The cutline is 70; voters at 10, 25, 40, 50 and 65 prefer 50, those at 80 and 95 prefer 90: 5 to 2.

Chapter 8 source: section "Median voter theorem".

Demonstration 4 of 4

Rights plus unanimity can cycle

Can personal rights and the Pareto rule both be respected?

Each right forces one comparison and unanimity forces a third. When L cares about what P reads, the three close into a cycle; when L is indifferent about P's reading, the Pareto link goes away.

Equation, written in LaTeX: a\succ_P b\succ_P c,

Equation, written in LaTeX: b\succ_L c\succ_L a,

Equation, written in LaTeX: c\succ a\succ b\succ c,

Scroll sideways for the whole equation

In state a no one reads the book, in b person P reads it and in c person L reads it. P has the right to decide a versus b, L to decide a versus c. Pareto: if both prefer x to y, so does society.

Predict first. If the Pareto rule is switched off, does a transitive social order exist?

Your prediction

Choose an example

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Figure: Rights plus unanimity can cycle. Directed triangle of states a, b and c with 3 required comparisons; a cycle.
L's view of P reading: Cares: b > c > a, Apply the Pareto rule: On
Constructed example: the chapter's hypothetical disputed book (both views of L); switching the Pareto rule off is added for comparison.

Calculated values

Required comparisons
3
Cycle
yes
Consistent social order
none

P's right over a versus b gives a > b; L's right over a versus c gives c > a. Both P and L rank b above c, so Pareto adds b > c. That makes 1 + 1 + 1 = 3 required comparisons, and c > a > b > c is a cycle, so no transitive social order exists.

Worked steps

  1. P's right: a > b
  2. L's right: c > a
  3. Both P and L rank b above c, so Pareto adds b > c.
  4. 3 comparisons: cycle

Use the idea

When rights and unanimous preferences conflict, look at which preferences are about other people's private choices.

Where the conclusion applies

Two people, three states and strict social comparisons from each rule.

Check your understanding: With L indifferent between a and b, which social order is consistent?
c > a (L's right) and a > b (P's right); Pareto is silent on b versus c, so c > a > b.

Chapter 8 source: section "Sen's liberal paradox".